/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 Maximum field from a passing cha... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 6

Maximum field from a passing charge In a colliding beam storage ring an antiproton going east passes a proton going west, the distance of closest approach being \(10^{-10} \mathrm{~m}\). The kinetic energy of each particle in the lab frame is \(93 \mathrm{GeV}\), corresponding to \(\gamma=100\). In the rest frame of the proton, what is the maximum intensity of the electric field at the proton due to the charge on the antiproton? For about how long, approximately, does the field exceed half its maximum intensity?

Short Answer

Expert verified
The relative speed between the proton and antiproton is the speed of light. In the rest frame of the proton, the antiproton is also observed to be moving at the speed of light. The maximum field can be determined using the formula \[E = \frac{1}{4\pi\epsilon_0} \frac{e}{(r/\gamma)^2} \]. The field remains above its half-maximum for a duration approximately the time it takes for the antiproton to travel the total length of the field distribution, which can be calculated by dividing \(2r\) by \(c\)

Step by step solution

01

Relative Speed in Lab Frame

First, determine the relative speed between proton and antiproton in the laboratory frame. Since both are moving in opposite directions, the addition (rather than subtraction) of their individual speeds applies per the principles of physics. Given that both the proton and the antiproton have a kinetic energy of 93 GeV and hence an associated speed very close to the speed of light, \(c\), the relative speed will also approximate \(c\).
02

Speed Transform

Next, using the principle of relativistic addition of velocities, transform this speed into the proton's rest frame to estimate how fast the proton sees the antiproton moving. Given both particles move at speed \(c\) in laboratory frame, their relative speed as observed in the proton's rest frame, \(v\), is calculated by: \[v = \frac{2c}{1+\frac{c^2}{c^2}} = c\] Hence, also in the proton's rest frame, the antiproton is seen to be moving at the speed of light.
03

Electric Field Calculation

The next step is to compute the maximum electric field. The electric field of a moving charge is given by: \[E = \frac{1}{4\pi\epsilon_0} \frac{e}{(r/\gamma)^2} \] Where e is the charge, \(\epsilon_0\) is the vacuum permittivity, r is the distance of closest approach. Substituting known values and \(gamma=100\) as given, calculate the maximum field.
04

Compute Half-Field Time

To estimate the time for which the electric field remains above its half maximum, consider the total length of the electric field distribution. Approximate it by considering the field to be non-zero within the range of \(-r\) to \(r\) with respect to the proton. As the antiproton is moving with the speed of light, the time during which the field surpasses half its maximum will approximately be the time it takes for the antiproton to travel the field's length, \(2r\), with speed \(c\). Therefore, the time can be calculated by conducting the division of \(2r\) divided by \(c\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Relativistic Velocity
When particles travel at speeds close to the speed of light, their velocities must be calculated using relativistic physics. This is because time and space start to behave differently at these extreme speeds. The velocity of one particle relative to another, especially when both are near light speed, involves more than simple addition or subtraction.
The relativistic velocity addition formula is used in such scenarios to accurately predict how fast one observer sees another object moving. The formula is:
  • \( v = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}} \)
For protons and antiprotons in a collider, each moving at \(c\), the resulting relative velocity remains close to \(c\). This ensures their interactions are considered in the relativistic framework.
Kinetic Energy
Kinetic energy represents the energy of motion. In particle physics, understanding a particle's kinetic energy is crucial, particularly when they approach relativistic speeds.
This energy can be described using the relation:
  • \( KE = (\gamma-1)mc^2 \)
where \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \) is the Lorentz factor. When \( KE \) reaches values as high as 93 GeV, as in the problem, \( \gamma \) becomes significant, showing just how much the moving mass increases in relation to the rest mass. This increase affects all calculations related to particle interactions and fields.
Proton-Antiproton Collision
In the scenario where a proton and antiproton collide, their interactions are incredibly significant. They are composed of quarks and antiquarks, leading to fascinating dynamics upon collision.
When these particles collide head-on at high energy levels, such as 93 GeV, they can create a range of different particles from the energy released. This energy is derived from their kinetic energy and relativistic nature, contributing to the field and collision outcomes.
The special characteristic of proton-antiproton collisions is that they are matter-antimatter interactions, which may result in annihilation, transforming mass into energy.
Maximum Field Intensity
Determining the electric field's maximum intensity is critical in understanding the interactions between high-speed particles. The electric field due to a moving charged particle is intensified by the factor \( \gamma \), given by:
  • \( E = \frac{1}{4\pi\epsilon_0} \frac{e}{(r/\gamma)^2} \)
Here, \( \gamma \) reduces the effective range \( r \), increasing field strength. Suppose the distance of closest approach is \(10^{-10}\) meters and \( \gamma = 100 \). In this case, the electric field is much stronger than for non-relativistic speeds.
The time during which the field exceeds half its intensity involves figuring out the particle's speed across the field region, affected by the relativistic speed and proximity-related factors.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Electron in an oscilloscope * The deflection plates in a high-voltage cathode ray oscilloscope ure two rectangular plates, \(4 \mathrm{~cm}\) long and \(1.5 \mathrm{~cm}\) wide, and spaced, \(.8 \mathrm{~cm}\) apart. There is a difference in potential of \(6000 \mathrm{~V}\) between he plates. An electron that has been accelerated through a potenial difference of \(250 \mathrm{kV}\) enters this deflector from the left, moving, of find the position of the electron and its direction of motion when t leaves the deflecting field at the other end of the plates. We shall lates is uniform right up to the end. The rest energy of the electron nay be taken as \(500 \mathrm{keV}\). (a) First carry out the analysis in the lab frame by answering the following questions: \- What are the values of \(\gamma\) and \(\beta ?\) \- What is \(p_{x}\) in units of \(m c ?\) \- How long does the electron spend between the plates? (Neglect the change in horizontal velocity discussed in Exercise 5.25.) \- What is the transverse momentum component acquired, in units of \(m c\) ? \- What is the transverse velocity at exit? \- What is the vertical position at exit? \- What is the direction of flight at exit?(b) Now describe this whole process as it would appear in an inertial frame that moved with the electron at the moment it entered the deflecting region. What do the plates look like? What is the field between them? What happens to the electron in this coordinate system? Your main object in this exercise is to convince yourself that the two descriptions are completely consistent.

Stationary rod and moving charge A charge \(q\) moves with speed \(v\) parallel to a long rod with linear charge density \(\lambda\), as shown in Fig. 5.30. The rod is at rest. If the charge \(q\) is a distance \(r\) from the rod, the force on it is simply \(F=q E=q \lambda / 2 \pi r \epsilon_{0}\) Now consider the setup in the frame that moves along with the charge \(q\). What is the force on the charge \(q\) in this new frame? Solve this by: (a) transforming the force from the old frame to the new frame, without caring about what causes the force in the new frame; (b) calculating the electric force in the new frame.

Maximum energy storage between cylinders ** We want to design a cylindrical vacuum capacitor, with a given radius \(a\) for the outer cylindrical shell, that will be able to store the greatest amount of electrical energy per unit length, subject to the constraint that the electric field strength at the surface of the inner cylinder may not exceed \(E_{0}\). What radius \(b\) should be chosen for the inner cylindrical conductor, and how much energy can be stored per unit length?

A charge inside a shell * Is the following reasoning correct or incorrect (if incorrect, state the error). A point charge \(q\) lies at an off-center position inside a conducting spherical shell. The surface of the conductor is at constant potential, so, by the uniqueness theorem, the potential is constant inside. The field inside is therefore zero, so the charge experiences no force.

A stationary proton is located on the \(z\) axis at \(z=a .\) A negative muon is moving with speed \(0.8 c\) along the \(x\) axis. Consider the total electric field of these two particles, in this frame, at the time when the muon passes through the origin. What are the values at that instant of \(E_{x}\) and \(E_{z}\) at the point \((a, 0,0)\) on the \(x\) axis?
See all solutions

Recommended explanations on Physics Textbooks

Astrophysics

Read Explanation

Oscillations

Read Explanation

Radiation

Read Explanation

Wave Optics

Read Explanation

Fluids

Read Explanation

Rotational Dynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.